JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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Application +4 | -1

\sum\limits_{{n=1}}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}

, then

\sum\limits_{{n=1}}^{\infty} \frac{1}{(2 n-1)^2}

View answer and explanation

Correct option: A

\frac{\pi^2}{8}

Option A (

\frac{\pi^2}{8}

) is correct because it is derived by subtracting the sum of the reciprocals of even squares from the sum of the reciprocals of all squares.

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